Question: Find the limit as $x$ approaches positive infinity. $\lim_{x\to\infty}\dfrac{2x^4-7}{\sqrt{4x^8+7x^5}}=$
Explanation: Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the numerator is $x^4$, let's divide by $x^4$. In the denominator, let's divide by $\sqrt{x^8}$, since for any value, $x^4=\sqrt{x^8}$. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{2x^4-7}{\sqrt{4x^8+7x^5}} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{2x^4-7}{x^4}}{\dfrac{\sqrt{4x^8+7x^5}}{\sqrt{x^8}}} \gray{\text{Divide sides by }x^4=\sqrt{x^8}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to\infty}\dfrac{\dfrac{2\cancel{x^4}}{\cancel{x^4}}-\dfrac{7}{x^4}}{\sqrt{\dfrac{4\cancel{x^8}}{\cancel{x^8}}+\dfrac{7\cancel{x^5}}{\cancel{x^5}\cdot x^3}}} \\\\ &=\lim_{x\to\infty}\dfrac{2-\dfrac{7}{x^4}}{\sqrt{4+\dfrac{7}{x^3}}} \\\\ &=\lim_{x\to\infty}\dfrac{2-0}{\sqrt{4+0}} \gray{\lim_{x\to\infty}\dfrac{k}{x^n}=0} \\\\ &=\dfrac{2}{\sqrt{4}} \\\\ &=\dfrac{2}{2} \\\\ &=1 \end{aligned}$ In conclusion, $\lim_{x\to\infty}\dfrac{2x^4-7}{\sqrt{4x^8+7x^5}}=1$.